mensuration area volumes Model Questions & Answers, Practice Test for ssc chsl tier 1 2023

Answer: (b)

r = 5 cm l = 13 cm

h = $√{l^2 - r^2} = √{13^2 - 5^2} = √{144}$ = 12

Hence, volume of cone = $1/3 × π × 4^2 × h$

= $1/3 × π × (5)^2 × 12 = 100 π$

Answer: (d)

Here, $AD^2$ = BD.DC

⇒ ${AD}/{BD} = {DC}/{AD}$

Hence, ΔABC must be right angled triangle.

So, $BC^2 = AB^2 + AC^2$

Answer: (a)

x = ∠VTR = 52°

x + z = 180°

(since, PTMR is a cyclic quadrilateral)

⇒ 52° + z = 180°

⇒ z = 128°

In ΔPTR, PT = TR

(given)

x = ∠1 = 52°

∠PTU = ∠1 = 52° ⇒ ∠QTU = y + 52°

⇒ 90° = y + 52° ⇒ y = 38°

∴ x + y + z = 52° + 38° + 128° = 218°

Answer: (d)

Let radius of cone be x and slant height of cone be 3x

Total surface area of cone = πr(r + l)

= π × x (x + 3x) = $4x^2$π

Curved surface area of cone = πrl = π × x × 3x = $3x^2$π

Required ratio = $4x^2π : 3x^2$π = 4 : 3

Answer: (c)

The perpendicular bisector of the chord of a circle always pass through the centre. So, Statement I is wrong

The angle in a semi-circle is a right angle. So, Statement II is correct.